Alternating Current (AC) – Transformer and AC Generator: NEET MCQs with Answers

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Syllabus covered: Transformer and AC Generator: AC generator and transformer

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Question 1

A simplified defibrillator must charge its capacitor to several thousand volts from the mains. The mains feeds a transformer, whose secondary feeds the capacitor through a diode. A two-way switch joins the capacitor either to the diode and secondary (position A) or to electrodes on the patient's chest (position B). What kind of transformer is needed, and what does the capacitor do in each switch position?
  1. Step-up; it stays charged in A and charges further in B.
  2. Step-down; it charges in A and discharges through the patient in B.
  3. Step-up; it charges in A and discharges through the patient in B.
  4. One-to-one; it discharges through the patient in both positions.
Show answer
Answer: (C) Step-up; it charges in A and discharges through the patient in B.
Explanation: The capacitor must store a large energy, which needs a voltage of thousands of volts, so the transformer must raise the mains voltage. In position A the capacitor is connected to the rectified supply and charges. In position B it is connected across the electrodes and its stored energy is released quickly through the patient.

Question 2

A plant generates power at 400 V and feeds a two-wire line of total resistance 25 Ω through an ideal step-up transformer. At the town, an ideal step-down transformer takes 200 kW from the line at a primary voltage of 5000 V. What voltage rating must the step-up transformer at the plant have?
  1. 400 V to 6000 V
  2. 400 V to 7000 V
  3. 400 V to 5000 V
  4. 400 V to 5500 V
Show answer
Answer: (A) 400 V to 6000 V
Explanation: The line current is 200 kW / 5000 V = 40 A. The voltage drop along the line is 40 × 25 = 1000 V. The sending end must therefore be at 5000 + 1000 = 6000 V, so the step-up transformer must raise 400 V to 6000 V.

Question 3

The armature of an AC generator rotates uniformly at 50 rad/s in a uniform magnetic field. The maximum flux linkage NΦ of the coil is 0.20 Wb. What is the magnitude of the induced emf at the instant when the flux linkage is 0.12 Wb?
  1. 4 V
  2. 10 V
  3. 6 V
  4. 8 V
Show answer
Answer: (D) 8 V
Explanation: The flux linkage is NΦ = 0.20 cos ωt and the emf is 0.20 ω sin ωt, so the peak emf is 0.20 × 50 = 10 V. When NΦ = 0.12 Wb, cos ωt = 0.6, which gives sin ωt = 0.8. The emf is then 10 × 0.8 = 8 V. Flux and emf are 90° out of phase, so the emf is not simply 60% of its peak.

Question 4

The charging unit of a defibrillator uses an ideal transformer with N_s/N_p = 10 on 220 V (rms) mains. The secondary charges a 40 μF capacitor through an ideal diode until the capacitor reaches the peak secondary voltage. What energy is stored in the capacitor?
  1. 1.94 J
  2. 387.2 J
  3. 193.6 J
  4. 96.8 J
Show answer
Answer: (C) 193.6 J
Explanation: The secondary rms voltage is 220 × 10 = 2200 V, so its peak value is 2200√2 V. The diode lets the capacitor charge to this peak and stops it discharging back into the winding. The stored energy is ½CV² = ½ × 40 × 10⁻⁶ × 2 × (2200)² = 193.6 J.

Question 5

An ideal autotransformer has one continuous winding of 1000 turns on an iron core, with 240 V (rms) applied across the whole winding. A load is connected across a tapped section of 250 turns of the same winding and draws 12 A (rms). Which statement is correct?
  1. The load voltage is 960 V, the input current is 3 A, and the load shares the winding with the supply.
  2. The load voltage is 60 V, the input current is 3 A, and the load shares the winding with the supply.
  3. The load voltage is 60 V, the input current is 3 A, and the load is electrically isolated from the supply.
  4. The load voltage is 60 V, the input current is 48 A, and the load is electrically isolated from the supply.
Show answer
Answer: (B) The load voltage is 60 V, the input current is 3 A, and the load shares the winding with the supply.
Explanation: Voltage divides in the ratio of turns, so the load voltage is 240 × 250/1000 = 60 V. For an ideal device input power equals output power, so 240 × I = 60 × 12 and I = 3 A. Because the primary and secondary are parts of one winding, the load is not electrically isolated from the supply.

Question 6

A generator coil of 200 turns and area 0.05 m² rotates at 2 revolutions per second in a uniform field of 0.04 T, about an axis perpendicular to the field. Which set correctly gives the peak emf, the average emf magnitude over the half-turn that starts from the position of maximum flux, and the average emf over one complete turn?
  1. 5.0 V, 3.2 V, zero
  2. 3.2 V, 5.0 V, zero
  3. 5.0 V, 2.5 V, 3.2 V
  4. 5.0 V, 3.2 V, 3.2 V
Show answer
Answer: (A) 5.0 V, 3.2 V, zero
Explanation: The peak emf is NBAω = 200 × 0.04 × 0.05 × 4π ≈ 5.0 V. In the half-turn the flux linkage goes from +NBA to −NBA, a change of 0.8 Wb in 0.25 s, so the average emf is 3.2 V, which equals 2/π of the peak. Over a full turn the flux linkage returns to its starting value, so the average emf is zero.

Question 7

A station sends 400 kW of power into a line of total resistance 5 Ω. What fraction of this power is lost as heat in the line if it is sent at 2 kV, and if it is sent at 20 kV?
  1. 5 % and 0.05 %
  2. 25 % and 2.5 %
  3. 50 % and 5 %
  4. 50 % and 0.5 %
Show answer
Answer: (D) 50 % and 0.5 %
Explanation: Line current is I = P/V, which is 200 A at 2 kV and 20 A at 20 kV. The losses I²R are 200² × 5 = 200 kW and 20² × 5 = 2 kW. These are 50 % and 0.5 % of 400 kW. Raising the voltage ten times reduces the loss a hundred times, which is why transformers step up the voltage for transmission.
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Question 8

Consider two statements about an AC generator. Statement I: The slip rings rotate with the armature coil, while the brushes pressing on them stay fixed. Statement II: The slip rings convert the alternating emf induced in the coil into a direct emf in the external circuit.
  1. Both statements are false
  2. Statement I is false and Statement II is true
  3. Statement I is true and Statement II is false
  4. Both statements are true
Show answer
Answer: (C) Statement I is true and Statement II is false
Explanation: The ends of the armature coil are joined to the slip rings, which turn with the shaft, and the fixed brushes slide on them. This lets current pass from the rotating coil to the stationary external circuit without twisting any wire. The rings do not rectify anything, so the current in the external circuit reverses every half revolution.

Question 9

Which pairing of an energy loss in a transformer with its method of reduction is NOT correct?
  1. Flux leakage: wind the primary and secondary far apart from each other
  2. Joule heating in the windings: use thick wires of low resistance
  3. Hysteresis in the core: use a material with a narrow hysteresis loop
  4. Eddy currents in the core: use a laminated core
Show answer
Answer: (A) Flux leakage: wind the primary and secondary far apart from each other
Explanation: Flux leakage occurs because some flux of the primary does not link the secondary. It is reduced by winding the two coils close together, often one over the other, so moving them apart makes it worse. Thick wires cut Joule heating, lamination raises the resistance to eddy currents, and a narrow-loop material wastes less energy per magnetising cycle.

Question 10

A town receives a fixed power over the same transmission line. The line voltage is raised from 11 kV to 33 kV using transformers. How do the heat loss in the line and the voltage drop along the line change?
  1. Heat loss becomes 1/3 and voltage drop becomes 1/9 of the earlier values
  2. Heat loss becomes 1/3 and voltage drop becomes 1/3 of the earlier values
  3. Heat loss becomes 1/9 and voltage drop becomes 1/3 of the earlier values
  4. Heat loss becomes 1/9 and voltage drop becomes 1/9 of the earlier values
Show answer
Answer: (C) Heat loss becomes 1/9 and voltage drop becomes 1/3 of the earlier values
Explanation: For the same power, I = P/V, so tripling the voltage reduces the line current to one-third. The voltage drop IR is proportional to I and falls to 1/3. The heat loss I²R is proportional to I² and falls to 1/9. This is why power is transmitted at high voltage.

Question 11

A sub-station draws 500 kW from a transmission line through the primary of an ideal step-down transformer, the primary voltage being 10 kV. The two-wire line has a total resistance of 8 Ω. Neglecting every other loss, how much power must the generating plant send into the line?
  1. 505 kW
  2. 520 kW
  3. 540 kW
  4. 500 kW
Show answer
Answer: (B) 520 kW
Explanation: The line current is I = P/V = 500 kW / 10 kV = 50 A. The heat produced in the line is I²R = 50² × 8 = 20 kW. The plant must therefore supply 500 kW + 20 kW = 520 kW.

Question 12

A power of 100 kW is sent through a line of total resistance 5 Ω at unity power factor. The power lost as heat in the line when the transmission voltage is 2 kV and when it is 20 kV is respectively
  1. 25 kW and 250 W
  2. 12.5 kW and 125 W
  3. 12.5 kW and 1.25 kW
  4. 2.5 kW and 25 W
Show answer
Answer: (B) 12.5 kW and 125 W
Explanation: The line current is I = P/V, which is 50 A at 2 kV and 5 A at 20 kV. The loss I²R is then 2500 × 5 = 12.5 kW and 25 × 5 = 125 W. Raising the voltage ten times reduces the current ten times and the line loss a hundred times.

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